Where <math>W(\omega_1,\omega_2,\omega_3,\omega_4)=2\cosh(K_1(\omega_1,\omega_2,\omega_3,\omega_4)){\text{exp}}[K_2(\omega_1\omega_4+\omega_2\omega_3)+K_1(\omega_1\omega_2+\omega_3\omega_4)] </math><br><br>

Where <math>W(\omega_1,\omega_2,\omega_3,\omega_4)=2\cosh(K_1(\omega_1,\omega_2,\omega_3,\omega_4)){\text{exp}}[K_2(\omega_1\omega_4+\omega_2\omega_3)+K_1(\omega_1\omega_2+\omega_3\omega_4)] </math><br><br>

−

Now impose another square lattice that crosses through the vertices of ''Figure 7'' diagonally (''Figure 8''). Just as was done for the original 16-vertex model, on each bond there is an associated arrow with two possible directions (this type representing Ising spins). Each vertex has the same 16 possible arrow (spin) configurations as shown before in ''Figure 4''. In sum, starting with an Ising model defined on a Kagomé lattice, we have recovered the 16-vertex model. By satisfying free-fermion conditions specific to the Kagomé lattice, free energy and critical conditions can be obtained, although specifics are again long and cumbersome and will not be written here.

+

Now impose another square lattice that crosses through the vertices of ''Figure 7'' diagonally (''Figure 8''). Just as was done for the original 16-vertex model, on each bond there is an associated arrow with two possible directions (this type representing Ising spins). Each vertex has the same 16 possible arrow (spin) configurations as shown before in ''Figure 4''. In sum, starting with an Ising model defined on a Kagomé lattice, we have recovered the 16-vertex model. By satisfying free-fermion conditions specific to the Kagomé lattice, free energy and critical conditions can be obtained, although specifics are again long and cumbersome and will not be written here.<br>

About me

My name is Dahlia and I am a first year grad student in chemistry. My specific program is called chemical physics. I just joined Eugene Shakhnovich's lab, where I plan to pursue theoretical biophysics.
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Fun facts on soft matter

Final Project: FRUSTRATED SYSTEMS AND SOFT MATTER

My project concerns the very complex subject of frustration. I hope that I will be able to introduce the topic to you with some very simple spin systems, and then illustrate why this is important to soft matter!

What is Frustration?

Frustration occurs in a myriad of different physical soft matter systems, ranging from polymer glasses, to proteins, to liquid crystals. Although the mathematics and physics of very simple frustrated systems could be shown in various ways, I believe that the clearest way to present it is through the use of magnetic spins. All sorts of hard and soft-condensed matter systems have polarization, whether it be of magnetic spins, electron spin, charge on a molecule, etc. The way spin frustration is dealt with analytically and computationally--on a lattice--is the same technique that is used in many protein folding models, polymer melt models, and really any other system for which you can figure out how to discretize it on a lattice. For the purpose of clarity, here I illustrate some of the simplest techniques with magnetic spin examples. Thus the math we cover here is much more generally applicable than simply to spin systems. It is not difficult to extend the underlying principles of frustrated spin systems to other frustrated systems in soft matter which I will discuss subsequently. Ian asked us to "stretch the boundaries of soft matter." Frustration is present in many, many soft-matter systems. It is a relatively advanced topic, and I think it nicely works to connect some of the physics we already know from hard condensed matter to the field of soft-condensed matter. I love connections like that. I'm assuming that everyone who reads this is familiar with the Ising model, but if you aren't there are lots of great statistical mechanics books that cover it in all its glory.

In its most general definition, when a system is in a frustrated state, it means that the minimum total energy of a system does not correspond to the sum of the minimum of all local interactions. Non-frustrated materials have a ground state that is characterized by a single potential well, which represents the uniform arrangement of perfectly ordered spins. By contrast, frustrated materials have a ground state that is an ‘energy landscape’ with many degenerate ground state configurations, separated by barriers of random height. This leads to low critical temperatures and strong dynamics at low temperatures. A simple example of a frustrated system is the following: In Figure 1, there is a triangular plaquette in which each spin must be opposite to one another. The blue and grey spins satisfy this easily, but the orange spin cannot. If its spin is down then it is not minimizing its interaction with the blue spin. If its spin is up the same happens with relation to the grey spin. The orange spin is thus frustrated. In other words, frustration occurs when a spin or multiple spins cannot find an orientation which fully satisfies all the interactions with its neighboring spins.

Figure 1

Before getting into more specifics, I first define some key features of frustrated systems. Two spins, Si and Sj with an interaction J have an interaction energy, E = -J (Si•Sj). If J > 0, the interaction is called ferromagnetic. Ferromagnetic materials have a minimum energy of –J, which occurs when all spins are parallel to one another. If J < 0, the interaction is called antiferromagnetic. Antiferromagnetic materials have a minimum energy of J, which occurs when all spins are anti-parallel to one another.

The type of frustration that emerges from the competing interactions on triangular, face-centered cubic, and hexagonal-close-packed lattice structures with antiferromagnetic nearest-neighbor interactions is called geometrical frustration. Figure 1 also illustrates geometrical frustration. Later I will discuss an analogous case in soft matter: a self-avoiding, random walking polymer with geometrical frustration.

In addition to geometrical frustration, there is also frustration that occurs due to competing interactions. For example,suppose there is a one dimensional system of spins which have the condition that nearest neighbor (nn) interactions are ferromagnetic and next nearest neighbor (nnn) interactions are anti-ferromagnetic. There is no physical way for all interactions to be minimized in this system, and hence it is frustrated. Below is a picture of this type of frustration:

There is also frustration in glassy systems (which we will discuss later in glassy polymers).
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Solving the Triangular System Exactly with nearest neighbor Interactions

Taking only into account nearest neighbor interactions, the ground state spin state of the triangle in picture A can be solved. First let Si (i=1,2,3) of amplitude S make an angle Ox axis. In this case, taking the energy to be the sum of the spin interactions:
<math>E=J(\mathbf{S_1}\cdot\mathbf{S_2}+\mathbf{S_2}\cdot\mathbf{S_3}+\mathbf{S_3}\cdot\mathbf{S_1})=JS^2\left[\cos(\theta_1-\theta_2+\cos(\theta_2-\theta_3\cos(\theta_3-\theta_1)\right] </math>
<math>\frac{\partial E}{\partial\theta}=-JS^2\left[\sin(\theta_1-\theta_2)-\sin(\theta_3-\theta_1)\right]=0 </math>
<math>\frac{\partial E}{\partial\theta}=-JS^2\left[\sin(\theta_2-\theta_3)-\sin(\theta_1-\theta_2)\right]=0 </math>
<math>\frac{\partial E}{\partial\theta}=-JS^2\left[\sin(\theta_2-\theta_3)-\sin(\theta_1-\theta_2)\right]=0 </math>
<math>\theta_1-\theta_2=\theta_2-\theta_3=\theta_3-\theta_1=\frac{2*\pi}{3}=120^o</math>
From this we can determine the ground state of the antiferromagnetic triangular lattice, depicted in picture below:

This is probably the simplest example of a frustrated system that can be solved analytically. However, all frustrated systems that are superimposed on a lattice follow a general scheme. Interactions between particles are taken into account by a model such as the Ising models. This allows us to find the ground state configuration of frustrated systems as well as important thermodynamic properties such as the free energy, specific heat, and the critical exponents. We can also find out where the phase transition (between the disordered and ordered state) of our material will occur. Of course, for all but a few cases, such terms are found via simulation (using Monte Carlo algorithms, Molecular Dynamics, etc).
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Solving frustrated Ising spin systems without disorder.

Systems without disorder have periodic interactions. A few of these systems are exactly solvable in 1 and 2 dimensions. A few such systems are: centered square lattice and generalized versions, the Kagomé lattice, an anisotropic centered honeycomb lattice. Like all frustrated systems, the ground states of these exactly solvable systems have very high, often infinite, degeneracy. Thermal fluctuations can preferentially select certain entropic states to others, thereby lowering levels of degeneracy. Another interesting property of these systems is that they have the property of the coexistence of order and disorder at equilibrium, which means that there will always be some spins that are disordered at all temperatures, even in an ordered phase.

The frustrated lattices can be described by using the 2D Ising model. Mathematically solving the Ising model involves first finding the partition function – a challenging problem. One commonly used method for finding the partition function involves mapping the Ising model onto a 16-vertex model (or a 32-vertext model). In this project I only describe the 16-vertex model and then show an example of how one can map the Ising model onto the 16-vertex model.

To understand the 16-vertex model, first imagine a lattice in which each vertex is connected by an edge (Figure 3). Each edge has two possible states, represented by arrows of different directions (up or down). Each vertex has 16 possible configurations, as shown in picture Figure 4. Each possible configuration has a different associated vertex weight (Boltzmann factor), <math>\omega_k = {\text{exp}}(\beta\epsilon_k)</math>, where <math>\epsilon_k</math> is the energy. The partition function is thus, <math>Z=\sum_c{\text{exp}}\left(-\beta\left[n_1\epsilon_1\right]\right)</math>

The sum is over all allowed configurations C of arrows on the lattice and <math>n_1</math> is the number of vertex arrangements of type j in a given configuration C. Z is a function of the Boltzmann weights: <math>Z=Z\left(\omega_1...\omega_{16}\right)</math>. In order to solve this exactly, the free-fermion condition has to be satisfied. In stipulating the free-fermion condition, the partition function is simplified. To satisfy the free-fermion condition, the following constraints are present,
<math>\omega_1=\omega_2</math>,<math>\omega_3=\omega_4</math>,<math>\omega_5=\omega_6</math>,<math>\omega_7=\omega_8</math>,<math>\omega_9=\omega_10</math>,<math>\omega_11=\omega_12</math>,<math>\omega_13=\omega_14</math>,<math>\omega_15=\omega_16</math>
and the free energy is expressed as
<math> f = \frac{-1}{4\pi\beta}\int_0^{2\pi}d\phi log(A(\phi)+Q(\phi)^\frac{1}{2})</math>
where:
<math> A(\Phi)=a+c\cos(\Phi)</math>
<math> Q(\Phi)=y^2+z^2-x^2 - 2yx\cos(\Phi)+x^2\cos^2(\Phi)</math>
<math> a=.5(\omega_1^2+\omega_3^2+2\omega_1\omega_3+\omega_5^2+\omega_7^2+2\omega_5\omega_7)+2(\omega_9^2+\omega_{13}^2))</math>
<math> c=2[\omega_9(\omega_1+\omega_3)-\omega_{13}(\omega_5+\omega_7)] </math>
<math> y=2[\omega_9(\omega_1+\omega_3)-\omega_{13}(\omega_5+\omega_7)] </math>
<math> z=.5[(\omega_1+\omega_3)^2-(\omega_5+\omega_7)^2]+2(\omega_9^2-\omega_{13}^2)]</math>
<math> x^2=z^2-.25[(\omega_1-\omega_3)^2-(\omega_5-\omega_7)^2]^2</math>

Phase transitions occur when one or more pairs of zeros of the expression <math>Q(\Phi)</math> close in on the real Φ axis and ‘pinch’ the path of integration in the expression of free energy. This occurs when <math>z^2=y^2</math>, or when <math>\omega_1+\omega_3+\omega_5+\omega_7+2\omega_9+2\omega_{13}=2{\text{max}}(\omega_1+\omega_3,\omega_5+\omega_7,2\omega_9,2\omega_{13}) </math>. This type of singularity in specific heat depends on whether </math>(\omega_1-\omega_3)^2-(\omega_5-\omega_7)^2\neq 0</math> (in which case there is a logarithmic singularity or <math>(\omega_1-\omega_3)^2-(\omega_5-\omega_7)^2=0</math> (in which case there is an inverse square root singularity). The mathematics required to figure this out is long and cumbersome and beyond the scope of this project.</br>
As an example of how a 2D Ising model is mapped onto the 16-vertex model, we examine the case of the Kagomé lattice (see Figure 5). An Ising model defined on a Kagomé lattice is shown in Figure 6. In this picture, single lines represent nn interactions <math>(J_1)</math> and double lines represent nnn interactions <math>(J_2)</math>. The hamiltonian is:
<math>H=-J_1\sum_{<ij>}\sigma_1\sigma_2 - J_2\sum_{<<ig>>}\sigma_1\sigma_2</math>
where the first sume is over nn and the second sum is over nnn.

To transform the system into an exactly solvable free-fermion model, one decimates the central spin of each elementary cell of the lattice. The result is the "checkerboard" lattice (Figure 7), which is analogous to the lattice depicted in Figure 3. Each shaded box in the checkerboard lattice has an associated Boltzmann weight <math>W(\omega_1,\omega_2,\omega_3,\omega_4)</math>. The partition function of this checkerboard Ising model is,
<math>Z=\sum_{\sigma}\Pi W(\omega_1,\omega_2,\omega_3,\omega_4)</math>
Where <math>W(\omega_1,\omega_2,\omega_3,\omega_4)=2\cosh(K_1(\omega_1,\omega_2,\omega_3,\omega_4)){\text{exp}}[K_2(\omega_1\omega_4+\omega_2\omega_3)+K_1(\omega_1\omega_2+\omega_3\omega_4)] </math>

Now impose another square lattice that crosses through the vertices of Figure 7 diagonally (Figure 8). Just as was done for the original 16-vertex model, on each bond there is an associated arrow with two possible directions (this type representing Ising spins). Each vertex has the same 16 possible arrow (spin) configurations as shown before in Figure 4. In sum, starting with an Ising model defined on a Kagomé lattice, we have recovered the 16-vertex model. By satisfying free-fermion conditions specific to the Kagomé lattice, free energy and critical conditions can be obtained, although specifics are again long and cumbersome and will not be written here.

Complex Phases of Matter: Spin Ice

We spent a significant amount of time in class discussing phases of matter. According to the introduction on the class wiki about this topic: "In the physical sciences, a phase is a set of states of a macroscopic physical system that have relatively uniform chemical composition and physical properties. A straightforward way to describe phase is "a state of matter which is chemically uniform, physically distinct, and (often) mechanically separable." Ice cubes floating on water are a clear example of two phases of water at equilibrium. In general, two different states of a system are in different phases if there is an abrupt change in their physical properties while transforming from one state to the other. Conversely, two states are in the same phase if they can be transformed into one another without any abrupt changes."

We also looked very closely at the phase diagram of water, shown below. The green dotted line denotes the liquid-solid phase boundary lines of water.

Water, however, is even more complex than this diagram demonstrates. There is a phase of water, or we could also term it more generally, a state of matter, called spin ice that reflects the frustration and disorder present in ice. Spin ices are frustrated ferromagnets. (How often do you think of ice as a ferromagnet?) In some ways we can think of the "ice-phase" as having a transition within itself in which it crosses the boundary of disorder-order. I am now going to go into more detail this very interesting description of ice. Note that more generally speaking, spin ice refers to the phase of matter that exists for any material made of up of a tetrahedron of ions (like water).

Pauling’s model of the hydrogen structure of water ice is a canonical model of disorder. In this model, there is zero point entropy, which means that at zero Kelvin there is residual entropy due to disorder. The system can never reach true thermal equilibrium. (Zero point entropy is not unique to water ice: Wannier and Anderson showed that Ising model antiferromagnets with triangular, tetrahedral, and other geometries also have zero point entropy). Spin ice is a microscopic mapping of Pauling’s water ice model onto a ferromagnet system. One key feature of spin ice is the Ice Rule. This rule states that for each oxygen, two protons (ie. hydrogen atoms) must be in the near position and two in the far position. This equates to 2 protons pointing inwards and 2 protons pointing outwards. See Figure 2 for clarification. Water itself has a very strong binding energy of 221 kilocalories per mole due to its strong covalent O-H bonds. These hydrogen atoms are much nearer to the oxygen than the other two hydrogen atoms (in the far position) which are (weakly) hydrogen bonded. This creates an electrostatic problem, as protons want to be as far away from each other as possible.

Figure 2

With this description of spin ice, we can now calculate for ourselves the zero point entropy approximation that Pauling calculated years ago. In 1 mole of ice, there are <math>NO^{2-}</math> ions. Each O-O line of contact has 2 possible proton positions, meaning that there are <math>2^{2N}</math> configurations. However, 10 of these are very unfavorable due to energetic instability: <math>OH_4^{2+}</math>, the <math>4</math> <math>OH_3^+</math>, the 4 <math>OH^-</math> and <math>O^{2-}</math>. Therefore, The number of possible configurations is <math>\Omega \leq 2^{2N}(\frac{6}{16})^N.</math> <math>S_0=k_b ln(\Omega)=Nk_bln(\frac{2}{2})=0.81\frac{Cal}{deg\cdot mol} </math> It turns out that this approximatin is accurate to 1%-2%.
The existence of zero point entropy raises the obvious question: isn’t the third law of thermodynamics (that there is zero entropy at zero Kelvin) being violated? The fact is that Pauling’s model assumes that protons remain in a disordered state, even at very low temperatures. This has been experimentally confirmed using neutron scattering experiments. Even more amazingly, at around 100 Kelvin, the proton structure of water ice ‘freezes’ and do not evolve over time over the manifold of possible states. Furthermore, there are no long range correlations, which implies that the sample behaves like a Gibbs ensemble of equivalent states. The ‘frozen’ spin arrangement that is observed cannot be the true ground state because of the existence of zero point entropy, but the question of why it is not zero still remains. Zero point entropy exists because the ‘frozen’ spins are not in thermal equilibrium with one another. Below 1 Kelvin, how does the system transition to a true ground state with zero entropy? This is an unanswered question! For now, zero point entropy is the ‘practical entropy’ of the system, and it appears that the 3rd law does not apply to systems on the time scale of any conceivable experiment. The expected entropy of the true ground state in thermal equilibrium is, of course, still zero.
Monte Carlo simulations demonstrate the properties of spin ice in both a zero and a finite applied magnetic field. In picture K, we see that as there is a build up of magnetic correlation, and a rather broad peak representing the specific heat of the phase transition. As the system cools there is a loss of magnetic entropy. The broad peak implies that there is not much long range magnetic order via thermodynamic phase transition. In b, there is evidence of Pauling zero point entropy. Change in entropy can be calculated using the formula,
<math>\partial S \equiv S(T_2)-S(T_1)= \int_{T_1}^{T_2} \frac{C_H(T)}{T} dT </math>

There are two possible spin orientations for each degree of freedom. The spin entropy in the high temperature paramagnetic regime is Rln(2) ≈ 5.76 J/mol•K. In this simulation we see that only about 3.9 J/mol•K are recovered. 5.76-3.9 =1.86 J/mol•K, approximately the round state entropy associated with the huge ground state deneracy of spin ice, So=(R/2)ln(3/2) = 1.68 J/mold•K. When there is applied magnetic field of 0.5 T, the spin reorientation barriers are lowered, meaning that entropy from more unfavorable states are included. Thus, in this case, ~1.7 J/mol•K are recovered.